Samuel Perales

Elementary not Easy: The Irrationality of Pi

15 min read

TLDR (click to show/hide)

Everyone knows that pi is irrational, but very few know why. You don't have to read both proofs in this article, but choosing to learn at least one of two will move you from that first camp into the (much cooler) second one. If proofs aren't your thing, you can instead scroll to the bottom of the page and at least learn a bit about a cool way of approximating pi.

Happy Pi day everyone! Just recently UT Math Club just had its first meeting of the semester starting with one of my favorite speakers; Tom Gannon. Tom is currently a 6th year grad student that I volunteered under for Sunday Morning Math Group a few years ago. He's a very engaging math speaker and does a great job of finding topics that are accessible to the entire audience.

His Math Club talk was on the Irrationality of Pi, a concept any math student is familiar with, but may not know the proof for. There are several interesting ways to prove this, so in this article, I want to show a few of a them that I enjoy, all of which use only basic calculus, but none of which are particularly easy to understand conceptually.

Niven's Proof

Ivan Niven wrote his proof on the Irrationality of Pi in 1947. Part of the reason this proof is elementary but not easy is because of the functions he defines. They are relatively easy to manipulate with basic calculus but don't make any sense until the proof is tied together right at the end. It is not a particularly intuitive proof.

He takes the definition of pi to be the smallest positive number for which $\sin \left(x\right)=0$ and assumes this solution is rational. That is, we suppose that $\pi =a/b$ for integers $a$ and $b$. We also define the polynomials

  $f\left(x\right)={\displaystyle \frac{\left(x^n\right(a-bx{)}^n)}{n!}}$

  $F\left(x\right)=f\left(x\right)-f^{\text{'}\text{'}}\left(x\right)+f^{(4)}\left(x\right)+\; ...\; +\left(-1{)}^n{f}^{\mathrm{(2n)}}\right(x)$

for any positive integer $n$ based on this $a,b$.

The idea is to prove the following two claims which will in turn give us a contradiction by the Fundamental Theorem of Calculus.

Claim 1: $F\left(0\right)+F\left(\pi \right)$ is an integer.

Claim 2: ${\int }_0^{\pi }$$f\left(x\right)\sin \left(x\right)dx=F\left(0\right)+F\left(\pi \right)$.

Having proved our two claims, we are essentially done. Since $f\left(x\right)$ and $\sin \left(x\right)$ are both positive for $0<x<\pi$, our integral is positive and so is the integer $F\left(0\right)+F\left(\pi \right)$.

From the bounds $f\left(x\right)<(\pi a{)}^n/n!$ and $\sin \left(x\right)<1$, we see that we can bound the size of our integral by any amount with large enough $n$. By choosing $n$ to be large enough such that the integral is bounded above by $1$, we see that $0<F\left(0\right)+F\left(\pi \right)<1$ which is our contradiction since there is no such integer between these values.

Laczkovich's Proof

Johann Lambert was the first to prove that pi is irrational in 1761. Since then, some simplificiations have been made to his proof. Here we'll look at Miklos Laczkovich's version of Lambert's proof.

First, we define the functions

$f_k\left(x\right)=1-x^2/k+x^4/2!k(k+1)+x^6/3!k(k+1)(k+2)+...$   $k\notin \{0,-1,-2,...\}$

Just as with Niven's Proof, we'll prove a few claims first and get a contradiction from there.

Claim 1: For all real values of $x$

$x^2{f}_{\mathrm{k+2}}\left(x\right)/k(k+1)=f_{\mathrm{k+1}}\left(x\right)-f_k\left(x\right)$

Proof of Claim 1: If we compare the coefficients of powers of $x$, we will see that they are the same on both sides.

Claim 2: For all real values of $x$, $\underset{\mathrm{k\; \to \; \infty }}{lim}$ $f_k\left(x\right)=1$.

Proof of Claim 2: $x^{\mathrm{2n}}/n!$ is bounded above by some $C$ since it converges to $0$ so

$\left|f_k\right(x)-1|\le \sum _{n=1}^{\mathrm{\infty }}C/k^n=C/(k-1)$

so as $k$ tends to infinity, we can add $1$ to both sides to see the result.

Claim 3: If $x\ne 0$ and $x^2\in \mathbb{Q}$,

$f_k\left(x\right)\ne 0$ and $f_{k+1}\left(x\right)/f_k\left(x\right)\notin \mathbb{Q}$   $\forall k\in \mathbb{Q}\backslash \{0,-1,-2,...\}$

Proof of Claim 3: First, we will prove the following lemma:

Lemma: Assume otherwise (that is, either $f_k\left(x\right)=0$, or $f_{k+1}\left(x\right)/f_k\left(x\right)\in \mathbb{Q}$, or both). Then there must be some $y\ne 0$ and $a,b\in \mathbb{Z}$ such that $f_k\left(x\right)=ay$ and $f_{k+1}\left(x\right)=by$

Proof of Lemma: There are two cases we can split this into:

Case 1: $f_k\left(x\right)=0$

In this case, choose $y=f_{k+1}\left(x\right),a=0,$ and $b=1$. Then we satisfy the lemma's conclusion and $y\ne 0$ since otherwise we would contradict claim 2.

Case 2: $f_k\left(x\right)\ne 0$

In this case, since $f_{k+1}\left(x\right)/f_k\left(x\right)\in \mathbb{Q}$, we can choose $a,b\in \mathbb{Z}$ such that $f_{k+1}\left(x\right)/f_k\left(x\right)=b/a$ and $y=f_k\left(x\right)/a=f_{k+1}\left(x\right)/b$. Again we satisfy the lemma's conclusion and $y\ne 0$ since otherwise we would contradict claim 2.

Now we choose $c\in \mathbb{N}$ such that $bc/k,ck/x^2,c/x^2\in \mathbb{Z}$. Consider the following sequence of functions:

$g_n=\{\begin{array}{ll}{f}_k(x)& \text{if }n=0\\ {\displaystyle \frac{c^n}{k(k+1)\cdots (k+n-1)}}{f}_{k+n}(x)& \text{otherwise}\end{array}$

We have that $g_0=f_k\left(x\right)=ay\in \mathbb{Z}y$ and $g_1=c{f}_{k+1}\left(x\right)/k=bcy/k\in \mathbb{Z}y$. Then by claim 1:

  $g_{n+2}={\displaystyle \frac{c^n}{k(k+1)\cdots (k+n-1)}}{\displaystyle \frac{x^2}{(k+n)(k+n+1)}}{f}_{k+n+2}(x)=...=\left({\displaystyle \frac{ck}{x^2}+\frac{cn}{x^2}}\right)g_{n+1}-{\displaystyle \frac{c^2}{x^2}}{g}_n$

So $g_{n+2}$ is a sum of multiples of $g_n$ and $g_{n+1}$ and in turn is also a multiple of $y$. Claim 2 gives us that each $g_n>0$ and since it is a multiple of $y$, it follows that $g_n>\left|y\right|$. But this is a contradiction of claim 1 since a sequence bounded below by some positive number cannot converge to $0$. So we have proven claim 3.

Finally, notice that $f_{1/2}(x)=\cos \left(2x\right)$ and $f_{3/2}(x)=\sin \left(2x\right)/2x$.

Since $f_{1/2}(\pi /4)=0$, claim 3 tells us that ${\pi }^2/16$ is irrational and therefore, $\pi$ is irrational.

A Fun Approximation: Buffon's Needle

The last bit of pi I'll talk about today is not a proof. Rather, its my favorite probabalistic approximation of the constant. Buffon's Needle has the following setup - For a needle of length $l$ and parallel lines distance $2l$ apart covering the plane, there is exactly a $1/\pi$ chance that the needle will cross one of the parallel lines when dropped at random.

I'll leave the calculus behind this one for you to explore, but its safe to say that it is significantly easier than these other two proofs. Instead, I'll just show you the nice little program I made that simulates this setup and outputs the current approximation of pi.

Resources

• Another Nice Explanation of Niven's Proof
• Generalized Buffon's Needle
• Buffon's Noodle